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Nearby freely falling particles can accelerate relative to one another even though each follows a geodesic. A local inertial frame removes the common gravitational acceleration at one event; it cannot remove the variation across an extended laboratory. That variation is a tidal force, measured covariantly by geodesic deviation.

Let $u^{\mu }$ be the reference four-velocity and ${\xi }^{\mu }$ an infinitesimal separation to a neighboring geodesic, with matching affine parameters. With the convention $[{\nabla }_{\beta },{\nabla }_{\gamma }]V^{\mu }=R^{\mu }{}_{\nu \beta \gamma }{V}^{\nu }$,This relates measured relative acceleration to the Riemann curvature tensor. Reversing the curvature-tensor convention reverses the displayed curvature sign, without changing the physical relative motion.

In a small freely falling frame outside a spherical mass, the weak-field tidal tensor has radial and transverse components proportional to $2,-1,-1$. ThusAn initially spherical cloud stretches radially and compresses in both transverse directions. Two separated test masses, with their changing separation measured by light, form the basic idea of a gravity gradiometer; the common falling motion drops out. A uniform apparent acceleration in an accelerated laboratory can occur even in flat Minkowski spacetime, so it is the separation-dependent tidal response that diagnoses curvature.

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.

The metric tensor and its inverse are diagonal: $g_{xx}=1$, $g_{yy}=f^2$, $g^{xx}=1$, $g^{yy}=f^{-2}$ on a patch where $f\ne 0$. Only ${\partial }_x{g}_{yy}=2f{f}^{\prime }$ is nonzero. The Christoffel symbol formula therefore giveswith all other components zero. Using the stated Riemann curvature tensor convention,so $\boxed{R^x{}_{yxy}=-f{f}^{\prime \prime }}$. This is the curvature of a two-dimensional warped metric; its Gaussian curvature is $-f^{\prime \prime }/f$.

For the Reissner-Nordstrom metric with $F=1-2M/r+Q^2/r^2$, the Kretschmann scalar isThe first expression follows by squaring the independent Riemann curvature tensor components in an orthonormal frame. This Kretschmann scalar diverges at $r=0$ for any nonzero $M$ or $Q$, but is finite at every positive root of $F$. Regular Eddington-Finkelstein coordinates show that those positive roots are coordinate singularities of the static chart. In the extremal case $Q^2=M^2>0$, its horizon value is $8/M^4$.

A Ricci-flat manifold has vanishing Ricci tensor. This is weaker than vanishing Riemann curvature tensor: the latter describes complete local flatness. A Kerr black hole supplies a Lorentzian manifold that is Ricci flat in its vacuum region but has nonzero Riemann curvature tensor.

The algebraic symmetries of the Riemann curvature tensor leave only one independent curvature component in two dimensions and forceConsequently the Einstein tensor vanishes identically.

A Riemannian metric is a smoothly varying positive-definite inner product $g_p$ on each tangent space $T_{p}M$ of a smooth manifold. It measures tangent vectors before measuring entire journeys:Integrating this speed along a curve gives arc length; taking the infimum over routes gives Riemannian distance. A smooth manifold together with this measuring rule is a Riemannian manifold.

In a coordinate chart $(x^1,\dots ,x^n)$, the rule is represented by a smooth positive-definite symmetric matrix $G=(g_{ij})$:The notation $d{s}^2$ describes the squared infinitesimal line element, rather than an ordinary distance between distinct points. Changing coordinates changes the matrix entries; it preserves the measured norm of the same tangent vector.

For the plane with $d{s}^2=4d{x}^2+d{y}^2$, the vector $(1,0)$ has norm $2$ and $(0,1)$ has norm $1$. The coordinate change $X=2x,Y=y$ turns this into $d{s}^2=d{X}^2+d{Y}^2$: coefficients that look different can describe the same geometry. On a circular cylinder the induced metric $R^{2}d{\theta }^2+d{z}^2$ similarly becomes Euclidean when unrolled. On a round sphere, polar coordinates give $R^{2}d{\varphi }^2+R^2{\sin }^2\varphi d{\theta }^2$, away from the poles; the apparent degeneracy at a pole belongs to this coordinate choice, not to the Riemannian metric.

The point of the construction is that measurements make sense intrinsically, even without a chosen surrounding Euclidean space. It gives a common starting point for geodesics, curvature encoded by the Riemann curvature tensor, and the Laplace-Beltrami operator. ^[ref].

For dimension $n\ge 3$, the Schouten tensor packages the Ricci tensor and scalar curvature in the trace part of the Riemann curvature tensor. In dimension three it is $S_{ij}=R_{ij}-R{g}_{ij}/4$, and the full Riemann curvature tensor is reconstructed from it.

The strong equivalence principle extends universality of free fall to bodies with appreciable gravitational self-energy and requires local experiments, including gravitational ones, in a freely falling laboratory to be independent of the laboratory's location and velocity. This statement neglects external tidal variation across the laboratory; it does not remove Riemann curvature tensor. In general relativity, universal metric coupling, fixed local gravitational coupling and the absence of extra preferred-frame gravitational fields implement this principle.

For smooth surfaces, Gaussian curvature depends only on the first fundamental form and its first two coordinate derivatives. Consequently any smooth local isometry $f:M\to N$ satisfies $K_N(f(p))=K_M(p)$. The surprise is that $K$ is defined using the shape operator, which describes bending in space, yet its value can be recovered from measurements made entirely within the surface.

In his 1827 paper, Carl Friedrich Gauss expressed Gaussian curvature through the metric coefficients and their partial derivatives in Article 11, then stated its invariance under development in Article 12. Article 13 explicitly compares a plane with surfaces developable onto it, including cylindrical surfaces. The paper was presented on 8 October 1827 and printed in 1828; the Morehead–Hiltebeitel English translation appeared in 1902. ^[ref]

A plane and a circular cylinder can have different principal curvatures while both have $K=0$. Their matching first fundamental forms exhibit the actual local isometry. In contrast, no nonempty open patch of a sphere of radius $R$ can be flattened smoothly into a plane while preserving its first fundamental form: $1/R^2$ cannot become zero. Making the patch small reduces practical distortion, but never produces an exact local isometry. This is a geometric obstruction to a perfectly distance-preserving flat map of a spherical world, independent of where one places a seam.

The following proof outline connects the two descriptions of Gaussian curvature. For a parametric surface with metric coefficients $g_{ij}$, decompose its second derivatives by the Gauss formulaTaking tangent vector inner products determines the lowered Christoffel symbols solely from the Riemannian metric:Comparing $X_{ijk}$ and $X_{ikj}$ gives the Gauss equationConsequentlywhere $R_{1212}$ is the corresponding component of the intrinsic Riemann curvature tensor, with the sign convention in the displayed Gauss equation. The final expression uses only $g_{ij}$ and its first two derivatives.

In dimension three the Riemann curvature tensor is $R_{ijpq}=g_{ip}{S}_{jq}-g_{iq}{S}_{jp}-g_{jp}{S}_{iq}+g_{jq}{S}_{ip}$, where $S$ is the Schouten tensor. Consequently every three-dimensional Ricci-flat Riemannian manifold is a flat Riemannian manifold. The implication fails in higher dimensions.

A vacuum spacetime has vanishing matter stress-energy tensor. In $D>2$ dimensions, tracing the Vacuum Einstein equations gives $R=2D\Lambda /(D-2)$ and hence $R_{ab}=2\Lambda g_{ab}/(D-2)$; in four dimensions this reduces to $R_{ab}=\Lambda g_{ab}$. With zero cosmological constant it is Ricci flat, although its full Riemann curvature tensor can be nonzero: a vacuum Kerr black hole and vacuum gravitational waves illustrate that distinction.

Use signature $(-,+,+,+)$, $x^0=ct$ and $R^{\rho }{}_{\sigma \mu \nu }={\partial }_{\mu }{\Gamma }^{\rho }{}_{\nu \sigma }-{\partial }_{\nu }{\Gamma }^{\rho }{}_{\mu \sigma }+\cdots$. For a weak stationary metric, the Christoffel symbol satisfies ${\Gamma }^i{}_{00}=-\frac{1}{2}{\partial }_i{g}_{00}$ to first order. Differentiation gives the displayed Riemann curvature tensor component; products of connections are second order. For $g_{00}=-1+2GM/(c^{2}r)$ it is $GM({\delta }_{ij}-3x_i{x}_j/r^2)/(c^2{r}^3)$. Its radial eigenvalue is negative and its transverse eigenvalues positive, giving radial stretching and transverse compression in the geodesic deviation equation with a positive curvature term on the left.

The Weyl tensor is the completely trace-free part of the Riemann curvature tensor. In four dimensions,